, compared to $ in haskell

Question

Possible duplicate:
Haskell: the difference between the two. (dot) and $ (dollar sign)

Ok, I understand that this is:

f(g(x)) 

can be rewritten:

 f $ g(x) 

and you can also rewrite:

 f . g(x) 

What I do not fully understand is where the two do not overlap in functionality. I conceptually understand that they are not completely overlapping, but can anyone clarify this for me once and for all?

Answer 1

 Prelude> :t ($) ($) :: (a -> b) -> a -> b Prelude> :t (.) (.) :: (b -> c) -> (a -> b) -> a -> c 

$ applies the function to the value. . consists of two functions.

So, I can write f $ gx , which "will apply f to (g of x)" or f . g $ x f . g $ x , which "applies the composition of f and g to x". One common style is to accumulate points to the left with dollar trailing. The reason is that f $ g $ x means the same as f . g $ x f . g $ x , but the expression f $ g itself is often meaningless (in fact, perhaps a type error), and the expression f . g f . g means "composition of f and g"

Answer 2

In addition to what has already been said, you need to use $ as a “glue application application” in the following cases:

 map ($3) [(4+),(5-),(6*),(+4),(*5),(^6)] --[7,2,18,7,15,729] 

None (.3) and (3) will work in this example.

Answer 3

"f $ gx" cannot be rewritten to "f. gx". In fact, the compiler will not even accept the second function, because "(.)" Has the type "(b → c) → (a → b) → (a → c)". I. The second argument must be a function, but "gx" is a value, not a function.